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Difference between revisions of "2007 USAMO Problems/Problem 5"

(I removed the "Solution 1" heading from the middle of the solution.~~~~)
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Let <math>\displaystyle{a_{n}}</math> be <math>7^{7^{n}}+1</math>. The result holds for <math>\displaystyle{n=0}</math> because <math>\displaystyle{a_0 = 2^3}</math> is the product of <math>\displaystyle{3}</math> primes.
 
Let <math>\displaystyle{a_{n}}</math> be <math>7^{7^{n}}+1</math>. The result holds for <math>\displaystyle{n=0}</math> because <math>\displaystyle{a_0 = 2^3}</math> is the product of <math>\displaystyle{3}</math> primes.
  
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=== Solution 1 ===
 
Now we assume the result holds for <math>\displaystyle{n}</math>. Note that <math>\displaystyle{a_{n}}</math> satisfies the [[recursion]]
 
Now we assume the result holds for <math>\displaystyle{n}</math>. Note that <math>\displaystyle{a_{n}}</math> satisfies the [[recursion]]
  
 
<div style="text-align:center;"><math>\displaystyle{a_{n+1}= (a_{n}-1)^{7}+1} = a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)</math></div>
 
<div style="text-align:center;"><math>\displaystyle{a_{n+1}= (a_{n}-1)^{7}+1} = a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)</math></div>
  
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Since <math>\displaystyle{a_n - 1}</math> is an odd power of <math>\displaystyle{7}</math>, <math>\displaystyle{7(a_n-1)}</math> is a perfect square. Therefore <math>\displaystyle{a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}</math> is a difference of squares and thus composite, i.e. it is divisible by <math>\displaystyle{2}</math> primes. By assumption, <math>\displaystyle{a_n}</math> is divisible by <math>\displaystyle{2n + 3}</math> primes. Thus <math>\displaystyle{a_{n+1}}</math> is divisible by <math>\displaystyle{2+ (2n + 3) = 2(n+1) + 3}</math> primes as desired.
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Since <math>\displaystyle{a_n - 1}</math> is an odd power of <math>\displaystyle{7}</math>, <math>\displaystyle{7(a_n-1)}</math> is a perfect square. Therefore <math>\displaystyle{a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}</math> is a difference of squares and thus [[composite]], i.e. it is divisible by <math>\displaystyle{2}</math> primes. By assumption, <math>\displaystyle{a_n}</math> is divisible by <math>\displaystyle{2n + 3}</math> primes. Thus <math>\displaystyle{a_{n+1}}</math> is divisible by <math>\displaystyle{2+ (2n + 3) = 2(n+1) + 3}</math> primes as desired.
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=== Solution 2 ===
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Notice that <math>7^{{7}^{k+1}}+1=(7+1) \frac{7^{7}+1}{7+1} \cdot \frac{7^{7^2} + 1}{7^{7^1} + 1} \cdots  \frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}</math>. Therefore it suffices to show that <math>\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}</math> is composite.
 +
 
 +
Let <math>x=7^{7^{k}}</math>. The expression becomes
 +
 
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<div style="text-align:center;"><math>\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1} = \frac{x^7 + 1}{x + 1}</math></div>
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which is the shortened form of the [[geometric series]] <math>\displaystyle x^{6}-x^{5}+x^{4}-x^3 + x^2 - x+1</math>. This can be [[factor]]ed as <math>\displaystyle (x^{3}+3x^{2}+3x+1)^{2}-7x(x^{2}+x+1)^{2}</math>
 +
 
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Since <math>x</math> is an odd power of <math>7</math>, <math>\displaystyle 7x</math> is a [[perfect square]], and so we can factor this by difference of squares. Therefore, it is composite.
  
 
== See also ==
 
== See also ==

Revision as of 17:35, 2 May 2007

Problem

Prove that for every nonnegative integer $n$, the number $7^{7^n}+1$ is the product of at least $\displaystyle 2n+3$ (not necessarily distinct) primes.

Solution

We prove the result by induction.

Let $\displaystyle{a_{n}}$ be $7^{7^{n}}+1$. The result holds for $\displaystyle{n=0}$ because $\displaystyle{a_0 = 2^3}$ is the product of $\displaystyle{3}$ primes.

Solution 1

Now we assume the result holds for $\displaystyle{n}$. Note that $\displaystyle{a_{n}}$ satisfies the recursion

$\displaystyle{a_{n+1}= (a_{n}-1)^{7}+1} = a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)$

Since $\displaystyle{a_n - 1}$ is an odd power of $\displaystyle{7}$, $\displaystyle{7(a_n-1)}$ is a perfect square. Therefore $\displaystyle{a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}$ is a difference of squares and thus composite, i.e. it is divisible by $\displaystyle{2}$ primes. By assumption, $\displaystyle{a_n}$ is divisible by $\displaystyle{2n + 3}$ primes. Thus $\displaystyle{a_{n+1}}$ is divisible by $\displaystyle{2+ (2n + 3) = 2(n+1) + 3}$ primes as desired.

Solution 2

Notice that $7^{{7}^{k+1}}+1=(7+1) \frac{7^{7}+1}{7+1} \cdot \frac{7^{7^2} + 1}{7^{7^1} + 1} \cdots \frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}$. Therefore it suffices to show that $\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}$ is composite.

Let $x=7^{7^{k}}$. The expression becomes

$\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1} = \frac{x^7 + 1}{x + 1}$

which is the shortened form of the geometric series $\displaystyle x^{6}-x^{5}+x^{4}-x^3 + x^2 - x+1$. This can be factored as $\displaystyle (x^{3}+3x^{2}+3x+1)^{2}-7x(x^{2}+x+1)^{2}$

Since $x$ is an odd power of $7$, $\displaystyle 7x$ is a perfect square, and so we can factor this by difference of squares. Therefore, it is composite.

See also

2007 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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